A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows: $P = \frac{a^3

Question

A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows: $P = \frac{a^3 b^2}{\sqrt{c} d}$. The percentage errors of measurement in $a, b, c$ and $d$ are $1\%$, $3\%$, $4\%$ and $2\%$ respectively. What is the percentage error in the quantity $P$?

Accepted Answer

13% Explanation: The relative error in $P$ is given by the sum of the relative errors of its components multiplied by their respective powers: $\frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}$. Substituting the given percentage errors: Percentage error in $P = (3 	imes 1\%) + (2 	imes 3\%) + (0.5 	imes 4\%) + (1 	imes 2\%) = 3\% + 6\% + 2\% + 2\% = 13\%$. Note that errors are always added to find the maximum possible error, even if the variable is in the denominator.

Options

Explanation